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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
<li><a href="sec_2-intro.html" data-scroll="sec_2-intro" class="internal">Linear and Nonlinear Equation</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
<li><a href="sec2_1.html" data-scroll="sec2_1" class="internal">Linear Equations</a></li>
<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
<li><a href="sec3_4.html" data-scroll="sec3_4" class="internal">Complex roots of the characteristic equations</a></li>
<li><a href="sec3_5.html" data-scroll="sec3_5" class="internal">Repeated Roots: Reduction of Order</a></li>
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<li><a href="sec4_4.html" data-scroll="sec4_4" class="internal">The Method of Variation of Parameters</a></li>
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<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="internal">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
<li><a href="sec6_1.html" data-scroll="sec6_1" class="internal">Introduction <span class="process-math">\(\&amp;\)</span> Basic Theory</a></li>
<li><a href="sec6_2.html" data-scroll="sec6_2" class="internal">Homogeneous System with Constant Coefficients</a></li>
<li><a href="sec6_3.html" data-scroll="sec6_3" class="internal">Complex Eigenvalues</a></li>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
<li><a href="sec7_2.html" data-scroll="sec7_2" class="internal">Eigenvalue Problems</a></li>
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<li><a href="sec7_4.html" data-scroll="sec7_4" class="internal">The Fourier Convergence Theorem</a></li>
<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<a href="ch_eight.html" data-scroll="ch_eight" class="internal"><span class="codenumber">8</span> <span class="title">Laplace transform</span></a><ul>
<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
<li><a href="sec8_2.html" data-scroll="sec8_2" class="internal">Finding Laplace Transforms</a></li>
<li><a href="sec8_3.html" data-scroll="sec8_3" class="internal">Finding inverse transforms using partial fractions</a></li>
<li><a href="sec8_4.html" data-scroll="sec8_4" class="internal">Solving ODEs and ODE Systems</a></li>
<li><a href="sec8_5.html" data-scroll="sec8_5" class="internal">Step input and Impulse problems</a></li>
<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec_4-intro"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">1.4</span> <span class="title">Motivation</span>
</h2>
<p id="p-10">Suppose we know the value of the quantity now and we wish to predict its value in the future. This quantity can be, for example, the temperature of coffee in a cup, the number of people infected with a virus, the concentration of carbon dioxide in the atmosphere. To do the prediction, we must know how quickly these quantities are changing. Mathematically, the rate of change of one quantity is the derivative. And practically the rate of change of a quantity will depend on the quantity itself. Therefore, we may model the problem as <span class="process-math">\(\frac{\textrm{d} y}{\textrm{d} t}=f(t, y)\text{.}\)</span> This equation contains derivative, so it is a differential equation. Through this equation and the initial condition, we may know the behavior of this quantity at any time.</p>
<p id="p-11">As one example, suppose that <span class="process-math">\(N(t)\)</span> is the number of bacteria growing on a plate of nutrients. At the start of the experiment, suppose that there are <span class="process-math">\(1000\)</span> bacteria, so <span class="process-math">\(N(0) = 1000\text{.}\)</span> The rate of change of <span class="process-math">\(N\)</span> will be proportional to <span class="process-math">\(N\)</span> itself: if there are twice as many bacteria, then N will grow twice as rapidly. So we have:</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq1_7">
\begin{equation}
\textrm{d}N/\textrm{d}t = \sigma N.\tag{1.4.1}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(\sigma\)</span> is a constant, and <span class="process-math">\(\textrm{d}N/\textrm{d}t\)</span> is the derivative (rate of change) of <span class="process-math">\(N\)</span> with respect to time. We would have to do further experiments to find out the value of <span class="process-math">\(\sigma\text{.}\)</span> We can easily verify that</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq1_8">
\begin{equation}
N(t) = 1000e^{\sigma t}\tag{1.4.2}
\end{equation}
</div>
<p class="continuation">is a solution of this differential equation with the given initial condition. To do this, first calculate <span class="process-math">\(N(0)\)</span> and verify that it is the same as the number given:</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq1_9">
\begin{equation}
N(0) = 1000 e^{0} = 1000.\tag{1.4.3}
\end{equation}
</div>
<p class="continuation">Next, calculate <span class="process-math">\(\textrm{d}N/\textrm{d}t\)</span> and verify that it satisfies the differential equation:</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq1_10">
\begin{equation}
\frac{\textrm{d} N} {\textrm{d} t }= 1000\sigma e^{\sigma t }= \sigma 1000e^{\sigma t}  = \sigma N \textrm{d} t\tag{1.4.4}
\end{equation}
</div>
<p class="continuation">as required.</p>
<p id="p-12">As another example, Suppose a 25 year-old plans to set aside a fixed amount every year of his/her working life, invests at a real return of <span class="process-math">\(6\%\text{,}\)</span> and retires at age 65. How much must he/she invest each year to have <span class="process-math">\(8,000,000\)</span> at retirement? Let <span class="process-math">\(S(t)\)</span> be the value of the investment at time <span class="process-math">\(t\text{,}\)</span> and let <span class="process-math">\(r\)</span> be the annual interest rate compounded after every time interval <span class="process-math">\(\Delta t\text{.}\)</span> We can also include deposits and let <span class="process-math">\(k\)</span> be the annual deposit amount, and suppose that an installment is deposited after every time interval <span class="process-math">\(\Delta t\text{.}\)</span> The value of the investment at the time <span class="process-math">\(t+ \Delta t\)</span> is then given by</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\label{eq1_7}
S(t+ \Delta t) = S(t)+(r \Delta t)S(t) +k \Delta t
\end{equation*}
</div>
<p class="continuation">where at the end of the time interval <span class="process-math">\(\Delta t\text{,}\)</span> <span class="process-math">\(r \Delta t S(t)\)</span> is the amount of interest credited and <span class="process-math">\(k \Delta t\)</span> is the amount of money deposited <span class="process-math">\((k&gt;0)\text{.}\)</span></p>
<p id="p-13">Rearranging (<a href="" class="xref" data-knowl="./knowl/eq1_7.html" title="Equation 1.4.1">(1.4.1)</a>), we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq1_7.html">
\begin{equation*}
\frac{S(t+\Delta t)-S(t)}{\Delta t}=r S(t)+k.
\end{equation*}
</div>
<p class="continuation">The equation for continuous compounding of interest and continuous deposits is obtained by taking the limit <span class="process-math">\(\Delta t \rightarrow 0\text{.}\)</span> The resulting differential equation is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq1_7.html">
\begin{equation*}
\frac{\mathrm{d}\, S}{\mathrm{d}\, t}=r S+k
\end{equation*}
</div>
<p class="continuation">which can be solved with the initial condition <span class="process-math">\(S(0)=S_0\text{,}\)</span> where <span class="process-math">\(S_0\)</span> is the initial capital.</p>
<p id="p-14">The solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq1_8.html">
\begin{equation*}
\label{eq1_8}
S=S_0 e^{rt}+\frac{k}{r} e^{rt}(1-e^{-rt}).
\end{equation*}
</div>
<p class="continuation">Suppose <span class="process-math">\(S_0=0\text{,}\)</span> then from (<a href="" class="xref" data-knowl="./knowl/eq1_8.html" title="Equation 1.4.2">(1.4.2)</a>), we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq1_8.html">
\begin{equation*}
k=\frac{rS(t)}{e^{rt}-1}.
\end{equation*}
</div>
<p class="continuation">Then according to the problem, we have <span class="process-math">\(r=0.06\)</span> and <span class="process-math">\(t=40\)</span> and <span class="process-math">\(S(t)=8,000,000\text{,}\)</span> therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq1_8.html">
\begin{equation*}
k=\frac{0.06 \times 8,000,000}{e^{0.06 \times 40}-1}=\$ 47,889~ \mathrm{year}^{-1}.
\end{equation*}
</div>
<p class="continuation">To save approximately eight million at retirement, the worker would need to save about 50,000 per year over his/her working life. Note that the amount saved over the worker’s life is approximately <span class="process-math">\(40 \times 50,000 =2,000,000\text{,}\)</span> while the amount earned on the investment (at the assumed <span class="process-math">\(6\%\)</span> real return) is approximately <span class="process-math">\(8,000,000-2,000,000=6,000,000\text{.}\)</span> The amount earned from the investment is about <span class="process-math">\(3 \times\)</span>the amount saved, even with the modest real return of <span class="process-math">\(6\%\text{.}\)</span> Sound investment planning is well worth the effort.</p></section></div></main>
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